Combustion Analysis- Empirical and molecular formula

[Dustyneo]

Homework Statement

Menthol, the substance we can smell
in mentholated cough drops, is composed of C, H, and O. A
0.1005-g sample of menthol is combusted, producing 0.2829 g
of CO₂ and 0.1159 g of H₂O. What is the empirical formula
for menthol? If menthol has a molar mass of 156 g/mol, what is its molecular formula?.
Book answer:
The empirical and molecular formulas are C ₁₀H₂₀O

2. Homework Equations
Whole-number multiple = molecular weight /empirical formula weight

The Attempt at a Solution

Cant arrive at the same answer. I started off using the masses of the combustion products, in the case of carbon

(0,2829 g CO₂ ) (mol CO₂ /44,0 g CO₂) (mol C/mol CO₂) =
0,00643 mol C
As for H and O, I got 0,129 mol and 0,00643 mol respectively. Then comparing the relative number of moles of each element

C: 0,00643/0,00643=1 H: 0,129/0,00643=20 O:0,00643/0,00643 = 1 lead me obtaining CH₂₀O as the empirical formula. What am I doing wrong?

 

[Chestermiller]

Please show us how you got the 0.129 mol for H.

 

[Borek]

And 0.00643 mol for O.

 

[Dustyneo]

Made a calculation mistake for H, the corrected value is 0,0129

(0,1159 g H₂O) (mol H₂O/ 18,0 g H₂O) (2 mol H/ mol H₂O)= 0,0129 mol H

For O, I also noticed that the actual number is closer to 0,00644:
(0,1159 g H₂O) (mol H₂O/ 18,0 g H₂O) ( mol O/ mol H₂O) = 0,00644 mol O

 

[mjc123]

How do you know that all the O in the H₂O, and none of that in the CO₂, comes from the O in the compound rather than the oxygen in the air? You need to take the equation
aC_xH_yO_z + bO₂ → cCO₂ + dH₂O
and balance it.

 

[Borek]

To add to what mjc123 wrote, finding mass of oxygen in the compound is quite easy once you know mass of carbon and hydrogen (and actually doesn't require stoichiometry).

 

[Dustyneo]

How do you know that all the O in the H₂O, and none of that in the CO₂, comes from the O in the compound rather than the oxygen in the air? You need to take the equation
aC_xH_yO_z + bO₂ → cCO₂ + dH₂O
and balance it.

Ok. In order to fill the values for a, c and d, I did these calculations,
Finding the amount of menthol molecules,

(0,1005 g C₁₀H₂₀O ) (mol C₁₀H₂₀O /156,09 G C₁₀H₂₀O) (6,02 X 10²³ C₁₀H₂₀O molecules / mol C₁₀H₂₀O ) = 3,88 x 10²⁰ menthol molecules

CO₂ molecules
(0,2829 g CO₂) (mol CO₂/44,0 g CO₂ ) (6,02 X 10²³ molecules CO₂ /mol CO₂) = 3,87 X 10²¹ CO₂ molecules
H₂O molecules
(0,1159 g H₂O) (mol H₂O /18,0 g)(6,02 x 10²³ water molecules/mol H₂O) = 3,88 x 10²¹ water molecules

Simplifying these number of molecules into the equation (number of water and CO₂ molecules is 10 times the amount of menthol molecules) I get

C₁₀H₂₀O + bO₂ → 10 CO₂ + 10 H₂O . Am I on the right track?

 

[mjc123]

That's cheating, because you are using the formula of menthol in your calculation.
Go back to what Borek said. You have 0.2829g CO₂. What mass of C does this contain? It can only have come from the menthol, so this is the mass of C in 0.1005g menthol.
You have 0.1159g water. How much H does this contain? It can only have come from the menthol, so this is the mass of H in 0.1005g menthol.
Now you have the mass of C and H, what is the mass of O in 0.1005g menthol?
Having the mass fractions, convert them into mole fractions, and work out the simplest whole-number ratio.

 

[Dustyneo]

What mass of C does this contain? It can only have come from the menthol
How much H does this contain? It can only have come from the menthol

can only have come from the menthol

I think I grasped where the mistake was . So I proceeded to only get the mass amount for C and H.
(0,2829 g CO₂)(mol CO₂/44,0 g CO₂)(mol C/mol CO₂ )(12,0 g C/mol C) = 0,0772 g C
(0,1159 g H₂O)(mol H₂O/18,0 g H₂O) (2 mol H/mol H₂O)( 1 g H/mol H) = 0,0129 g H
Then substracting the addition of these masses from the menthol to get the amount of O:
(0,1005 g) - (0,0772 + 0,0129) = 0,0104 g O
Obtaining moles and comparing relative numbers
(0,0772 g C)(mol C/ 12,0 g C) = 0,00643 mol C
(0,0129 g H)(mol H/ g H)= 0,0129 mol H
(0,0104 g O)(mol O/16,0 g O)= 0,0006504 mol H

C: 0,00643/0,0006504≈ 10 H: 0,0129/0,0006504≈20 O: 0,0006504/0,0006504=1 giving C₁₀H₂₀O as the empirical formula, with a formula weight of aprox. 156 amu. Then using the relevant equation
156 amu/156 amu=1 indicates C₁₀H₂₀O also being the molecular formula. Thanks for the help.